Bomb Calorimetry constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = –q rxn = q bomb + q water
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# Bomb Calorimetry

Feb 07, 2016

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Bomb Calorimetry. constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = –q rxn = q bomb + q water. Example 4. - PowerPoint PPT Presentation
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Bomb Calorimetry

constant volume often used for combustion reactions heat released by reaction is absorbed

by calorimeter contents need heat capacity of calorimeter qcal = –qrxn = qbomb + qwater

Example 4When 0.187 g of benzene, C6H6, is burned in a bomb calorimeter,

the surrounding water bath rises in temperature by 7.48ºC. Assuming that the bath contains 250.0 g of water and that the calorimeter has a heat capacity of 4.90 kJ/ºC, calculate the energy change for the combustion of benzene in kJ/g.

C6H6 (l) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l)

qcal = (4.90 kJ/ºC)(7.48ºC) = 36.7 kJ

qrxn = -qcal = -36.7 kJ / 0.187 g C6H6 = -196 kJ/g

qV = ∆E = -196 kJ/g

Energy and Enthalpy

Most physical and chemical changes take place at constant pressure

Heat transferred at constant P - enthalpy (H) Can only measure ∆H ∆H = Hfinal - Hinitial = qP

sign of ∆H indicates direction of heat transfer

system

heat

system

heat

∆H > 0endothermic

∆H < 0exothermic

Energy and Enthalpy

Enthalpy and Phase Changes

Enthalpy and Phase Changes Melting and freezing

Quantity of thermal energy that must be transferred to a solid to cause melting - heat of fusion (qfusion)

Quantity of thermal energy that must be transferred from a solid to cause freezing - heat of freezing (qfreezing)

qfusion = - qfreezing

heat of fusion of ice = 333 J/g at 0°C

Enthalpy and Phase Changes Vaporization and condensation

Similarly: qvaporization = - qcondensation

heat of vaporization of water = 2260 J/g at 100°C 333 J of heat can melt 1.00 g ice at 0°C but it will boil

only:

333 J x (1.00 g / 2260 J) = 0.147 g water

Enthalpy and Phase Changes

H2O (l) H2O (g)

H2O (s) H2O (l)

H2O (g) H2O (l)

H2O (l) H2O (s)

endothermic

exothermic

State Functions Value of a state function

is independent of path taken to get to state - depends only on present state of system

Internal energy is state function

State Functions q and w not state

functions

Enthalpy and PV Work

H - state function

q - not a state function

How do internal energy and enthalpy differ?E = q + w

H = q P

so how can ∆H = q??

Enthalpy and PV Work

Example 5A gas is confined to a cylinder under constant

atmospheric pressure. When the gas undergoes a particular chemical reaction, it releases 89 kJ of heat to its surroundings and does 36 kJ of PV work on its surroundings. What are the values of ∆H and ∆E for this process?

q = -89 kJ w = -36 kJ

@ constant pressure:

∆H = qP = -89 kJ

∆E = ∆H + w = -89 kJ - 36 kJ = -125 kJ

Thermochemical Equations

∆H = Hfinal - Hinitial = H(products) - H(reactants)

∆Hrxn - enthalpy or heat of reaction

2 H2 (g) + O2 (g) 2 H2O (l) ∆H° = -571.66 kJ

coefficients of equation represent # of moles of reactants and products producing this energy change

Thermochemical Equations

∆H° standard enthalpy change defined as enthalpy change at 1 bar

pressure and 25°C

“Rules” of Enthalpy Enthalpy is an extensive property - magnitude of ∆H depends

on amounts of reactants consumed

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ∆H = -890 kJ

2 CH4 (g) + 4 O2 (g) 2 CO2 (g) + 4 H2O (l) ∆H = -1780 kJ

Enthalpy change of a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reactionCO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g) ∆H = +890 kJ

Enthalpy change for a reaction depends on the states of the

reactants and productsH2O (l) H2O (g) ∆H = +44 kJ

H2O (s) H2O (g) ∆H = +50 kJ

Example 6

Hydrogen peroxide can decompose to water and oxygen by the reaction:

2 H2O2 (l) 2 H2O (l) + O2 (g) ∆H = -196 kJ

Calculate the value of q when 5.00 g of H2O2 (l) decomposes.

mol H2O2 = 5.00 g( )1 mol

34.0146 g⎛⎝⎜

⎞⎠⎟ = 0.147 mol H2O2

H = qP = 0.147 mol H2O2( )-196 kJ

2 mol H2O2

⎝⎜⎞

⎠⎟ = -14.4 kJ

Example 7

Consider the following reaction, which occurs at room temperature and pressure:

2 Cl (g) Cl2 (g) ∆H = -243.4 kJ

Which has the higher enthalpy under these conditions, 2 Cl or Cl2?

2 Cl (g)

Example 8

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates:

Ag+ (aq) + Cl- (aq) AgCl (s) ∆H = -65.5 kJ

(a) Calculate ∆H for the formation of 0.200 mol of AgCl by this reaction.

H = 0.200 mol AgCl( )-65.5 kJ

1 mol AgCl

⎛⎝⎜

⎞⎠⎟

= -13.1 kJ

Example 8 (cont’d) Ag+ (aq) + Cl- (aq) AgCl (s) ∆H = -65.5 kJ

(b) Calculate ∆H when 0.350 mmol AgCl dissolves in water.

AgCl (s) Ag+ (aq) + Cl- (aq) ∆H = +65.5 kJ

H = 0.350 x 10-3 mol AgCl( )+65.5 kJ

1 mol AgCl

⎛⎝⎜

⎞⎠⎟

= 0.0229 kJ = 22.9 J

Bond Enthalpies

during chemical reaction bonds are broken and made

breaking bonds requires energy input (endothermic)

formation of bonds releases energy (exothermic)

weaker bonds broken and stronger bonds formed

Hess’s Law we can calculate ∆H for a reaction using ∆Hs

for other known reactions ∆H is a state function - result is same no

matter how we arrive at the final state Hess’s Law - if a reaction is carried out in a

series of steps, ∆H for overall reaction is equal to sum of ∆Hs for steps

Hess’s LawC (s) + O2 (g) CO2 (g) ∆H = -393.5 kJ

CO (g) + 1/2 O2 (g) CO2 (g) ∆H = -283.0 kJ

What is ∆H for C (s) + 1/2 O2 (g) CO (g) ???

C (s) + O2 (g) CO2 (g) ∆H = -393.5 kJ

CO2 (g) CO (g) + 1/2 O2 (g) ∆H = +283.0 kJ

C (s) + 1/2 O2 (g) CO (g) ∆H = -110.5 kJ

Example 9Calculate ∆H for the conversion of S to SO3 given the following

equations:

S (s) + O2 (g) SO2 (g) ∆H = -296.8 kJ

SO2 (g) + 1/2 O2 (g) SO3 (g) ∆H = -98.9 kJ

want S (s) SO3 (g)

S (s) + O2 (g) SO2 (g) ∆H = -296.8 kJ

SO2 (g) + 1/2 O2 (g) SO3 (g) ∆H = -98.9 kJ

S (s) + 3/2 O2 (g) SO3 (g) ∆H = -395.7 kJ

Enthalpies of Formation tables of enthalpies (∆Hvap, ∆Hfus, etc.)

∆Hf - enthalpy of formation of a compound from its constituent

elements. magnitude of ∆H - condition dependent standard state - state of substance in pure form at 1 bar and

25°C

∆Hf ° - change in enthalpy for reaction that forms 1 mol of

compound from its elements (all in standard state)

∆Hf ° of most stable form of any element is 0

CO2:

C (graphite) + O2 (g) CO2 (g) ∆Hf ° = -393.5 kJ/mol

Calculating ∆Hrxn° from ∆Hf° we can use ∆Hf° values to calculate ∆Hrxn° for any reaction

∆Hrxn° = ∑ [n ∆Hf° (products)] - ∑ [n ∆Hf° (reactants)]

C6H6 (l) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l)

∆Hrxn° = [(6 mol)(-393.5 kJ/mol) + (3 mol)(-285.83 kJ/mol)]

- [(1 mol)(49.0 kJ /mol) + (15/2 mol)(0 kJ/mol)]

∆Hrxn° = -3267 kJ/mol

Example 10Styrene (C8H8), the precursor of polystyrene polymers, has a standard

heat of combustion of -4395.2 kJ/mol. Write a balanced equation for the combustion reaction and calculate ∆Hf ° for styrene (in

kJ/mol).

∆Hf ° (CO2) = -393.5 kJ/mol

∆Hf ° (H2O) = -285.8 kJ/mol

C8H8 (l) + 10 O2 (g) 8 CO2 (g) + 4 H2O (l)

∆Hrxn ° = -4395.2 kJ/mol = [(8 mol)(-393.5 kJ/mol) + (4)(-285.8)]

- [(1) ∆Hf ° (C8H8) + (10)

(0)]

∆Hf ° (C8H8) = 104.0 kJ/mol